82. The distance from the relaxed position of the bottom end of the spring to its equilibrium position when the body is attached is given by Hooke’s law: Δx= F/k= (0.20 kg)(9.8 m/s2)/(19 N/m) = 0.103 m. (a) The body, once released, will not only fall through the Δxdistance but continue through the equilibrium position to a “turning point” equally far on the other side. Thus,
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.